Incircles on a regular tetrahedron

Suppose that we are looking at the tetrahedron from the side, and that we are considering the incircles of the triangles $ACD$ and $BCD$ . Suppose that the tetrahedron has side length $2r$ and that the angle between the faces $ACD$ and $BCD$ is $2\theta$ . Then the altitude of each triangular face has length $r\sqrt{3}$ , the radius of each incircle is $\tfrac{r}{\sqrt{3}}$ and $\sin\theta = \tfrac{1}{\sqrt{3}}$ . Points on the two incircles have position vectors $\begin{aligned} \mathbf{x} & = \; \frac{r}{\sqrt{3}}\left[(1 + \cos(s+t))\left(\begin{array}{c} \sin\theta \\ 0 \\ \cos\theta \end{array}\right) + \sin(s+t)\left(\begin{array}{c}0 \\ 1 \\ 0 \end{array}\right) \right] \\[2ex] \mathbf{y} & = \; \frac{r}{\sqrt{3}}\left[(1 + \cos(s-t))\left(\begin{array}{c} -\sin\theta \\ 0 \\ \cos\theta \end{array}\right) + \sin(s-t)\left(\begin{array}{c}0 \\ 1 \\ 0 \end{array}\right) \right] \end{aligned}$

and so $\mathbf{x}-\mathbf{y} \; = \; \frac{r}{\sqrt{3}}\left(\begin{array}{c} \sin\theta(2 + \cos(s+t)+\cos(s-t)) \\ \sin(s+t)-\sin(s-t) \\ \cos\theta(\cos(s+t)-\cos(s-t))\end{array}\right) \; = \; \frac{2r}{\sqrt{3}}\left(\begin{array}{c}\sin\theta(1 + \cos s \cos t) \\ \cos s \sin t \\ -\cos\theta \sin s \sin t \end{array}\right)$

and hence $\begin{aligned} \Vert\mathbf{x}-\mathbf{y}\Vert^2 & = \; \tfrac43r^2\big[\sin^2\theta(1 + \cos s \cos t)^2 + \cos^2s \sin^2t + \cos^2\theta\sin^2s \sin^2t\big] \\ & = \; \tfrac43r^3\big[\sin^2\theta(1 + 2\cos s \cos t + \cos^2s) + \cos^2\theta\sin^2t\big] \; = \; \tfrac49r^2\big[1 + 2\cos s \cos t + \cos^2s + 2\sin^2t\big] \\ & = \; \tfrac49r^2\big((\cos s + \cos t)^2 + 3 - 3\cos^2t\big) \end{aligned}$ The maximum value of the function $(u+v)^2 + 3 - 3v^2 = 3+\tfrac32u^2 - \tfrac12(u-2v)^2$ over the range $-1 \le u,v \le 1$ is $\tfrac92$ , achieved when $u=\pm1$ and $v=\tfrac12u$ . Thus the maximum value of $\Vert\mathbf{x}-\mathbf{y}\Vert^2$ is $2r^2$ , making the maximum distance between two points on these incircles $r\sqrt{2}$ . In this case $r=50$ , so the distace is $50\sqrt{2}$ , making the answer $\boxed{52}$ . Note that this maximum distance is achieved when the two points on the incircles are at the midpoints of diametrically opposite edges of the tetrahedron.